What is the hydroxide ion concentration of a 0240 m oxalic...

What is the hydroxide-ion concentration of a 0.240 M oxalic acid, H2C2O4, solution? For oxalic acid, Ka1 = 5.6 × 10–2 and Ka2 = 5.1 × 10–5. A) 2.9 × 10–12 M C) 3.5 × 10–3 M E) 1.0 × 10–7 M B) 8.6 × 10–14 M D) 1.1 × 10–13 M 4. What is the equilibrium concentration of H2C2O4 in a 0.230 M oxalic acid, H2C2O4, solution? For oxalic acid, Ka1 = 5.6 × 10–2 and Ka2 = 5.1 × 10–5. A) 1.4 × 10–1 M C) 2.3 × 10–1 M B) 1.1 × 10–1 M D) 5.1 × 10–5 M 5. A solution of aniline (C6H5NH2, Kb = 4.2 × 10–10) has a pH of 8.75 at 25°C. What was the initial concentration of aniline? A) 1.8 × 10–9 M C) 7.7 × 10–2 M E) 5.6 × 10–6 M B) 9.3 × 10–9 M D) 4.2 × 10–10 M 6. Which of the following salts will produce an acidic solution when added to pure water? A) Cs2CO3 C) KCN E) Li2S B) NaF D) Al(NO3)3 7. Which of the following salts is most likely to form an aqueous solution having the pH shown in the figure? (The pH in the picture is 4.55) A) NH4Cl C) K2CO3 E) LiNO3 B) NaBr D) RbCN

Transcript

00:02 Based on the collagative property, let us calculate molality of the solution.

00:13 Depression in freezing point.

00:22 Depression in freezing point equal to k1 multiplied with molality of the solution.

00:34 M.

00:37 Depression is t delta t f f equal to kf multiplied with m.

00:54 M equal to, as we need to find out molality, let us solve it for molality delta tf divided by kf.

01:05 Delta tf values 3 .52 degrees centigrade divided by kf value is 5 .12 degrees centigrade k z per mole.

01:19 Molality of the solution m is equal to we got the value as molality of solution is equal to 0 .675m.

01:38 Now we have to determine total moles of solute.

01:43 To determine total moles of a solute, molality of solution equal to total moles of solute divided by.

02:09 By mass of solvent, mass of solvent, benzene in kg.

02:33 Total moles of solute is equals to, total modes of solute equal to, let us solve this for total modes molality of the solution that is m molality of the solution m multiplied with mass of the solvent of benzene mass of solvent of benzene in k g let us substitute the values here 0 .675 is the molality of the solution and mass of solvent of benzene is 0 .035kg.

03:29 Simplification gives us 0 .0246 moles.

03:37 Total moles of solute equal to 0 .0 .0 .2406 moles.

03:55 Now let us calculate the mass of naphthalene.

03:59 To calculate mass of naphthalene, let us take total moles of solute total moles of solute equal to moles of naphthaline plus moles of pyrine modes of pyrine this is total moles.

04:28 Let us substitute the values in this total mass of solute total mass of solute includes a value of 3 .40 grams this is the mass of mixture so this includes so when we say this is mass of a mixture it is obvious that we said moles of naphthaline and plus most of pyrine is equals to total modes of solute so based on this we can say this is mass of naphthaline and mass of pyrine, naphthalene and pyrine.

05:19 Let mass of naphthalene, mass of naphthalene in grams, is a grams, that is we need to find it out, so we are using a.

05:35 Mass of naphthalene is a grams.

05:38 Molar mass molar mass of naphthaline molar mass of naphthalene equal to 128 grams per mole moles of naphthaline moles of naphthaline in a mixture equal to mass of naphthaline in a mixture divided by molar mass of naphthaline molar mass of naphthalene which is equal to a divided by 128 modes now let us consider mass of pyrine mass of pyrine in the mixture is equal to 340 minus a that is the mass of pyrine molar mass of pyrin just as we have considered molar mass of naphthaline here we are considering molar mass of pyrine is 202 grams per mole.

07:32 Moles of pyrine in the mixture, it is equals to, let us include the value of 340 minus a divided by 202.

07:50 Total moles let us consider.

07:54 Total moles of solute...

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